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A particle of charge $q$ and mass $m$ is moving with a constant velocity v along $+x$ - axis. It enters a region of constant magnetic field along $-z$ - axis from $x = a$ to $x = b$. The minimum value of $ v$ required so that the particle can enter the region $x>b$ is

$\begin {array} {1 1} (a)\;\large\frac{qbB}{m} & \quad (b)\;\large\frac{qaB}{m} \\ (c)\;\large\frac{q(b+a)B}{m} & \quad (d)\;\large\frac{q(b-a)B}{m} \end {array}$


1 Answer

The particle cannot enter $x>b$ if the radius of the circular path is lesser than the width of the region.
The width of the region is (b-a).
we have v= r B q /m.
For minimum value of v ,r must be equal to (b-a)
Ans : (d)


answered Feb 19, 2014 by thanvigandhi_1
edited Sep 15, 2014 by thagee.vedartham

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